A study was conducted to investigate the effectiveness of a drug in reducing a person’s weight. Using the data below calculate the mean d and standard deviation sd of the differences.
A study was conducted to investigate the effectiveness of a drug in reducing a person’s weight. Using the data below calculate the mean d and standard deviation sd of the differences.
subject | pete | doug | lyn | fred | gill | hank |
before | 189.9 | 211.3 | 211.5 | 198.1 | 206.3 | 211.6 |
after | 187.4 | 208.5 | 208.1 | 198.0 | 203.2 | 206.1 |
difference | 2.5 | 2.8 | 3.4 | 0.1 | 3.1 | 5.5 |
Difference is given by : 2.5 2.8 3.4 0.1
Sum of difference =8.8
Mean of difference =8.8/4=2.2
Sum of square of difference =
2.5*2.5 + 2.8*2.8 + 3.4*3.4 + 0.1*0.1 = 25.66
Variance =
=
= 1/4*(25.66-4*2.2*2.2)
= 1.575
Standard deviation is positive square root of variance = squareroot(1.575)
=1.25499
Mean of difference is : 2.2
standard deviation of difference is: 1.25499
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